Axial Deformation Member - Theory of Nonlinear Structural Analysis - page 169

e

P

100 mm

δ = 3 or 6 mm

5 mm

P

l b = 4.9 m

Figure 5.6 Model of the steel tube in tension.

(2) 6 mm displacement input
Since the axial displacement δ = 6 mm is larger than the yielding displacement, δ y = 0.0048
m, the steel tube has stepped into plasticity. The axial inelastic displacement and axial force
can be solved through Eqs. (5.14) and (5.15) as

=1−0 : 1

δ 00 =1−ð Þ δ − δ y

ð Þ 0 : 006−0 : 0048 ð Þ=0 : 0011 m ð5 : 29Þ

+ P y =0 : 1×8 : 14 × 10 7 ×0 : 0011 + 389500 = 398454 N

P = α k b δ − δ y ð5 : 30Þ

Example 5.2 A steel tube in compression
Consider the steel tube is now subjected to −0.005 m axial displacement at the support as shown
in Figure 5.7. Let us investigate how to solve the axial force should be applied on the steel tube.

Solution
According to Eqs. (5.17) and (5.24), the buckling load and corresponding displacement are
obtained as

P b = −180210 N ð5 : 31Þ

δ b = −0 : 00265 m ð5 : 32Þ

and the axial force and displacement at intermediate Point A a based on Point A are

calculated as:

P b 0 = −90105 N ð5 : 33Þ

δ b 0 = −0 : 0013 m ð5 : 34Þ

Since the given displacement δ = −0.005 m at the support is less than the intermediate
displacement δ b = −0.0013 m, the steel tube has stepped into the nonlinear compressive region
A−B. By substituting Eq. (5.22) into Eq. (5.6), the axial load is obtained as:

P = −138400 N ð5 : 35Þ